1.

If 3.01xx10^(20) molecules of H_(2)SO_(4) are removed from 98mg of H_(2)SO_(4). Then number of moles of H_(2)SO_(4) left are:

Answer»

`0.1xx10^(-3)` mol
`0.5xx10^(-3)` mol
`1.66xx10^(-3)` mol
`9.95xx10^(-2)` mol

Solution :Removed number of moles of `H_(2)SO_(4)=(3.01xx10^(20))/(6.02xx10^(23))`
`=5xx10^(-4)`
`=0.5xx10^(-3)`
Initial number of moles of `H_(2)SO_(4)=(0.098)/(98)=0.001=10^(-3)`
REMAINING moles of `H_(2)SO_(4)=10^(-3)-0.5xx10^(-3)=0.5xx10^(-3)`


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